H(t)=t^2-8t-48

Solution for H(t)=t^2-8t-48 equation:

(H)=H^2-8H-48

We move all terms to the left:

(H)-(H^2-8H-48)=0

We get rid of parentheses

-H^2+H+8H+48=0

We add all the numbers together, and all the variables

-1H^2+9H+48=0

a = -1; b = 9; c = +48;
Δ = b2-4ac
Δ = 92-4·(-1)·48
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{273}}{2*-1}=\frac{-9-\sqrt{273}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{273}}{2*-1}=\frac{-9+\sqrt{273}}{-2}
$